Given a number N, the task is to find the smallest number whose square has N digits.
Examples:
Input: N = 2
Output: 4
Explanation:
32 = 9, which has 1 digit.
42 = 16, which has 2 digits.
Hence, 4 is the smallest number whose square has N digits.
Input: N = 3
Output: 10
Explanation:
102 = 100, which has 3 digits.
Naive Approach: The simplest approach to solve the problem is to calculate the square of each number starting from and count the number of digits in its square. Print the first number whose square is obtained to be of N digits.
Time Complexity: O(?(10N))
Efficient Approach: To solve the problem, we need to make the following observations:
The smallest number whose square has 1 digit = 1
The smallest number whose square has 2 digits = 4
The smallest number whose square has 3 digits = 10
The smallest number whose square has 4 digits = 32
The smallest number whose square has 5 digits = 100
Hence, these numbers form a series 1, 4, 10, 32, 100, 317, …….
Now, we need to find a formula for the Nth term of the series.
The terms of the series can be expressed in the following form:
If N = 1, Smallest number possible is 1.
If N = 2, Smallest number possible is 41.
If N = 3, Smallest number possible is 10.
Hence, we can conclude that the Nth of the series can be expressed as
Hence, in order to solve the problem, we just need to calculate ceil(10(N – 1)/ 2) for the given integer N.
Below is the implementation of the above approach:
C++
// C++ Program to find the smallest// number whose square has N digits#include <bits/stdc++.h>using namespace std;// Function to return smallest number// whose square has N digitsint smallestNum(int N){ // Calculate N-th term of the series float x = pow(10.0, (N - 1) / 2.0); return ceil(x);}// Driver Codeint main(){ int N = 4; cout << smallestNum(N); return 0;} |
Java
// Java program for above approach class GFG{ // Function to return smallest number// whose square has N digitsstatic int smallestNum(int N){ // Calculate N-th term of the series float x = (float)(Math.pow(10, (N - 1) / 2.0)); return (int)(Math.ceil(x));}// Driver code public static void main(String[] args) { int N = 4; System.out.print(smallestNum(N)); } } // This code is contributed by spp |
Python3
# Python3 Program to find the smallest# number whose square has N digitsimport math;# Function to return smallest number# whose square has N digitsdef smallestNum(N): # Calculate N-th term of the series x = pow(10.0, (N - 1) / 2.0); return math.ceil(x);# Driver CodeN = 4;print(smallestNum(N));# This code is contributed by Code_mech |
C#
// C# program for above approach using System;class GFG{ // Function to return smallest number// whose square has N digitsstatic int smallestNum(int N){ // Calculate N-th term of the series float x = (float)(Math.Pow(10, (N - 1) / 2.0)); return (int)(Math.Ceiling(x));}// Driver code public static void Main() { int N = 4; Console.Write(smallestNum(N)); } } // This code is contributed by Code_Mech |
Javascript
<script>// java script Program to find the smallest// number whose square has N digits// Function to return smallest number// whose square has N digitsfunction smallestNum(N){ // Calculate N-th term of the series x = Math.pow(10.0, (N - 1) / 2.0); return Math.ceil(x);}// Driver Codelet N = 4;document.write(smallestNum(N)); // This code is contributed by Gottumukkala Bobby</script> |
32
Time Complexity: O(log(N))
Auxiliary Space: O(1)
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